Electronic Journal of Differential Equations

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Electronic Journal of Differential Equations
Vol. 1994(1994), No. 04, pp. 1-10. Published July 8, 1994.
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp (login: ftp) 147.26.103.110 or 129.120.3.113
EXISTENCE RESULTS FOR NON-AUTONOMOUS
ELLIPTIC BOUNDARY VALUE PROBLEMS
V. Anuradha, S. Dickens, and R. Shivaji
Abstract. We study solutions to the boundary value problems
−∆u(x) = λf (x, u);
u(x) + α(x)
∂u(x)
= 0;
∂n
x∈Ω
x ∈ ∂Ω
where λ > 0, Ω is a bounded region in RN ; N ≥ 1 with smooth boundary ∂Ω,
α(x) ≥ 0, n is the outward unit normal, and f is a smooth function such that it has
either sublinear or restricted linear growth in u at infinity, uniformly in x. We also
consider f such that f (x, u)u ≤ 0 uniformly in x, when |u| is large. Without requiring
any sign condition on f (x, 0), thus allowing for both positone as well as semipositone
structure, we discuss the existence of at least three solutions for given λ ∈ (λn , λn+1 )
where λk is the k-th eigenvalue of −∆ subject to the above boundary conditions. In
particular, one of the solutions we obtain has non-zero positive part, while another
has non-zero negative part. We also discuss the existence of three solutions where
one of them is positive, while another is negative, for λ near λ1 , and for λ large when
f is sublinear. We use the method of sub-super solutions to establish our existence
results. We further discuss non-existence results for λ small.
1.
Introduction. We first consider the boundary value problem
−∆u(x) = λf (x, u);
u(x) + α(x)
∂u(x)
= 0;
∂n
x∈Ω
x ∈ ∂Ω
(1.1)
(1.2)
where ∆ is the Laplacian operator, Ω is a bounded region in RN ; N ≥ 1 with
smooth boundary ∂Ω, α(x) ≥ 0, n is the unit outward normal, and f is a smooth
function such that
f (x, u)
= σ uniformly in x,
(1.3)
lim
u→∞
u
Submitted: January 23, 1994.
1991 AMS SubjectClassification. 35J65.
Partially supported by NSF Grants DMS - 8905936 and DMS - 9215027.
This work was completed while R. Shivaji was a Visiting Research Scientist at Georgia Institute
of Technology in 1993.
c
1994
Southwest Texas State University and University of North Texas.
1
V. Anuradha, S. Dickens, and R. Shivaji
2
EJDE–1994/04
f (x, u)
= β uniformly in x,
u→−∞
u
(1.4)
∂f
(x, u) ≥ 0.
∂u
(1.5)
lim
and
Let λk , φk be the eigenvalues and corresponding eigenfunctions of the boundary
value problem
(1.6)
−∆φk = λk φk ; x ∈ Ω
φk (x) + α(x)
∂φk (x)
= 0;
∂n
x ∈ ∂Ω.
(1.7)
Let I = [a, b] ⊂ (λn , λn+1 ) and for λ ∈ I consider the unique solution Zλ of the
boundary value problem
−∆Zλ − λZλ = −1;
Zλ (x) + α(x)
x∈Ω
∂Zλ (x)
= 0.;
∂n
x ∈ ∂Ω
(1.8)
(1.9)
Let µ1 = inf x∈Ω̄, λ∈I Zλ (x), µ2 = supx∈Ω̄, λ∈I Zλ (x) and µ = max{|µ1 |, µ2 }. Note
that Zλ+ 6≡ 0 (see Appendix 1). Also, ∃ δ(Ω) > 0 such that if I ⊂ (λ1 , λ1 + δ) then
Zλ > 0; x ∈ Ω (anti-maximum principle: see Clémént and Peletier in [1]).
Now assume that
∃ m > 0 such that y + m > f (x, y) > y − m ∀ y ∈ [−bmµ, bmµ]
and
β, σ <
1
b||wα ||∞
(1.10)
(1.11)
where wα is the unique positive solution to
−∆wα = 1;
wα (x) + α(x)
x∈Ω
∂wα (x)
= 0;
∂n
(1.12)
x ∈ ∂Ω.
(1.13)
Then we prove:
Theorem 1.1. Let I = [a, b] ⊂ (λn , λn+1 ) and (1.3)–(1.5), (1.10)–(1.11) hold.
Then these exists at least three solutions to (1.1)–(1.2) for λ ∈ I. One of these
solutions is a non-negative or sign changing solution, while another is a non-positive
or sign changing solution.
Remark 1.1. Note that unlike in the literature of “jumping nonlinearities” (see [2]–
[3]), our results do not require (β, σ) to include a part of the spectrum of −∆,
nor require, a part of the spectrum of −∆ to lie between f 0 (0) and f 0 (±∞) for
autonomous nonlinearities as in [4] and the reference within.
EJDE–1994/04
Non-autonomous EllipticBoundary Value Problems
3
Theorem 1.2. Let I = [a, b] ⊂ (λ1 , λ1 + δ) and (1.3)–(1.5), (1.10)–(1.11) hold.
Then there exists at least three solutions to (1.1)–(1.2) for λ ∈ I, where one is a
positive solution while another is a negative solution.
Next we consider the particular case when σ = β = 0 (sublinear case). We
further assume that there exists f1 (u) < f (x, u); ∀ x ∈ Ω̄, u ≥ 0 such that
Z r1
0
f (s) ds > 0
(1.14)
f1 (r1 ) = 0, f1 (r1 ) < 0,
0
for some r1 > 0, and that there exists f2 (u) > f (x, u); ∀ x ∈ Ω̄, u ≤ 0 such that
Z 0
f2 (r2 ) = 0, f20 (r2 ) < 0,
f (s) ds < 0
(1.15)
r2
for some r2 < 0. Then we prove:
Theorem 1.3. Assume (1.14)–(1.15) and let (1.3)–(1.5) hold with σ = β = 0.
Then there exists at least three solutions to (1.1)–(1.2) for λ large, where one is a
positive solution while another is a negative solution.
Remark 1.2. We refer to [4] where a positive solution for λ ∈ I = [a, b] ⊂ (λ1 , λ1 +δ)
was discussed in the case of autonomous, sublinear (σ = 0), and semipositone
(f (0) < 0) problems with Dirichlet boundary conditions. In [5] it was assumed
that there exists m > 0 such that f (y) ≥ y − m ∀y ∈ [0, bmµ] to obtain a positive
solution for λ ∈ I. Further by constructing a function f1 (u) satisfying (1.14) a
positive solution for λ large was discussed. See also [6] where the authors study
the existence of a positive solution via degree theory arguments. Hence they allow
non-autonomous problems with Robin boundary condition, but consider only the
sublinear case, and study the existence of positive solutions for λ large with the
assumption f (x, 0) ≤ 0 ∀x ∈ Ω̄.
We also consider f such that
∃ r > 0 for which f (x, r) ≤ 0 while f (x, −r) ≥ 0 for every x ∈ Ω̄.
(1.16)
Then we prove:
Theorem 1.4. Let I = [a, b] ⊂ (λn , λn+1 ) and (1.10), (1.16) hold. Assume r ≥
bmµ. Then there exists at least three solutions to (1.1)–(1.2) for λ ∈ I. One
of these solutions is a non-negative or sign changing solution, while another is a
non-positive or sign changing solution.
Theorem 1.5. Let I = [a, b] ⊂ (λ1 , λ1 + δ) and (1.10), (1.16) hold. Assume
r ≥ bmµ. Then there exists at least three solutions to (1.1)–(1.2) for λ ∈ I. One
of these is a positive solution, while another is a negative solution.
Theorem 1.6. Let (1.16) hold, and assume that (1.14) holds for all 0 ≤ u ≤ r
with r1 ≤ r while (1.15) holds for all −r ≤ u ≤ 0 with r2 ≥ −r. Then there exists
at least three solutions to (1.1)–(1.2) for λ large, where one is a positive solution
while another is a negative solution.
Remark 1.3. Consider the boundary value problem
−∆u(x) = λf (||x||, u);
x ∈ BN
4
V. Anuradha, S. Dickens, and R. Shivaji
u(x) + α
∂u(x)
= 0;
∂n
EJDE–1994/04
x ∈ ∂BN
where α ≥ 0 is a constant and BN is the unit ball in RN . Then under the corresponding hypothesis on f (||x||, u) instead of f (x, u), one can generate all the
solutions obtained in Theorems 1.1–1.6 to be radial. This follows from the fact
that all the sub and super solutions we will use in the proofs of these theorems will
turn out to be radial.
We next discuss non-existence results under the assumption that there exists
γ > 0 such that
∂f
≤ γ.
(1.17)
∂u
We recall that λ1 is the principal eigenvalue and φ1 > 0 is a corresponding eigenfunction of −∆ subject to boundary Robin conditions (1.2). Then we prove:
R
Theorem 1.7. Assume Ω f (x, 0)φ1 (x) ≤ 0 and λ < λ1 /γ. If u(6≡ 0) is any
solution of (1.1)–(1.2) then u− 6≡ 0 (i.e. there does not exist any solution u(6≡ 0)
which is non-negative).
R
Theorem 1.8. Assume Ω f (x, 0)φ1 (x) ≥ 0 and λ < λ1 /γ. If u(6≡ 0) is any
solution of (1.1)–(1.2) then u+ 6≡ 0 (i.e. there does not exist any solution u(6≡ 0)
which is non-positive).
Theorem 1.9. Assume f (x, 0) ≤ 0 and λ < λ1 /γ. If u is any solution of (1.1)–
(1.2) then u ≤ 0.
Theorem 1.10. Assume f (x, 0) ≥ 0 and λ < λ1 /γ. If u is any solution of (1.1)–
(1.2) then u ≥ 0.
We give detailed proofs of our results in section 2. For literature on autonomous
semipositone problems with Dirichlet boundary conditions see [7], while for the
positone case see [8] and the references cited in [8].
Our existence results are based on sub-super solutions. Namely, a super solution
is defined as a smooth function φ such that
−∆φ ≥ λf (x, φ);
φ(x) + α(x)
∂φ(x)
≥ 0;
∂n
x∈Ω
x ∈ ∂Ω
(1.18)
(1.19)
and a subsolution is a smooth function ψ that satisfies (1.18)–(1.19) with the inequalities reversed. If ψ ≤ φ, then it follows that (1.1)–(1.2) has a solution u such
that ψ ≤ u ≤ φ (see [9]–[10]).
Further if ψ1 is a subsolution, ψ2 is a strict subsolution, φ1 is a strict supersolution and φ2 is a supersolution such that ψ1 ≤ ψ2 ≤ φ2 , ψ1 ≤ φ1 ≤ φ2 ,
ψ2 (x0 ) > φ1 (x0 ) for some x0 ∈ Ω̄, then (1.1)–(1.2) has at least three distinct
solutions u1 , u2 , u3 such that u1 ≤ u2 ≤ u3 . See [11] for this multiplicity result
which was proved for Dirichlet boundary conditions. However, it is easy to see that
this result holds for the Robin boundary conditions (1.2) as well.
EJDE–1994/04
2.
Non-autonomous EllipticBoundary Value Problems
5
Proofs of Theorems 1.1–1.10.
Proof of Theorem 1.1. Let v1 (x) := bmZλ (x) and u2 (x) = −v1 (x). Then
−∆v1 = bm(−∆Zλ ) = bm(λZλ − 1) ≤ bm(λZλ − λ/b)( since λ ≤ b) = λ[bmZλ − m]
= λ[v1 − m] < λf (x, v1 ) by (1.10), and −∆u2 = ∆v1 ≥ λ[−v1 + m] = λ[u2 + m]
> λf (x, u2 ) again by (1.10). Thus v1 is a strict subsolution while u2 is a strict
supersolution. Note that v1+ 6≡ 0 and u−
2 6≡ 0. Now let u1 (x) := Jwα (x) where
J > 0 is large enough so that
f (x, J||wα ||∞ )
1
≥
b||wα ||∞
J||wα ||∞
(2.1)
and
u1 ≥ v1 ,
u1 ≥ u2 .
(2.2)
Here (2.1) is possible since σ < ||wα1||∞ b , and (2.2) is possible by the Hopf’s maximum principle. Then −∆u1 = J ≥ λf (x, J||wα ||∞ ); x ∈ Ω (by (2.1) since λ ≤ b)
˜ α (x) where J˜ > 0
≥ λf (x, Jwα ) (since ∂f
≥ 0) = λf (x, u1 ). Next let v2 (x) = −Jw
∂u
is large enough so that
˜ α ||∞ )
1
f (x, −J||w
≥
˜ α ||∞
b||wα ||∞
−J||w
(2.3)
and
v2 ≤ v1 ,
v2 ≤ u2 .
(2.4)
1
||wα ||∞ b
and (2.4) is possible by the Hopf’s
˜
˜ α ||∞ ); x ∈ Ω (by (2.3)
maximum principle. Then −∆v2 = −J ≤ λf (x, −J||w
∂f
˜ α ) (since
since λ ≤ b) ≤ λf (x, −Jw
∂u ≥ 0) = λf (x, v2 ). Thus u1 is a supersolution
while v2 is a subsolution such that v2 ≤ v1 ≤ u1 and v2 ≤ u2 ≤ u1 , where v1 is
a strict subsolution, u2 is a strict supersolution with v2 ≤ 0, v1+ 6≡ 0, u−
2 6≡ 0 and
u1 ≥ 0. Hence the result. Here again (2.3) is possible since β <
Proof of Theorem 1.2. If λ ∈ (λ1 , λ1 + δ) then Zλ > 0; x ∈ Ω. Then v1 (x) > 0;
x ∈ Ω while u2 < 0; x ∈ Ω and hence the result follows by the proof in Theorem 1.1. Proof of Theorem 1.3. Consider the autonomous Dirichlet problem
−∆v = λf1 (v);
v = 0;
x∈Ω
x ∈ ∂Ω.
(2.5)
(2.6)
Then by (1.14) it follows that there exists λ̄1 > 0 such that for λ ≥ λ̄1 (2.5)λ
(2.6) has a positive solution vλ (see [12]). Clearly since ∂v
∂n ≤ 0; x ∈ ∂Ω and
f1 (S) < f (x, S); ∀x ∈ Ω̄, S ≥ 0 vλ is a strict subsolution to (1.1)–(1.2) for λ ≥ λ̄1 .
Next consider
(2.7)
−∆u = λf2 (u); x ∈ Ω
u = 0;
x ∈ ∂Ω.
(2.8)
6
V. Anuradha, S. Dickens, and R. Shivaji
EJDE–1994/04
Then setting w = −u we see that w satisfies
−∆w = λ[−f2 (−w)];
w = 0;
x∈Ω
x ∈ ∂Ω.
(2.9)
(2.10)
R −r2
Let g(w) = −f2 (−w). Then g(−r2 ) = 0, g0 (−r2 ) = f20 (r2 ) < 0 and 0 g(s)ds
R0
R −r2
= 0 −f2 (−s)ds = − γ2 f2 (s)ds > 0. Hence again by [12], there exists λ̄2 > 0
such that for λ ≥ λ¯2 , (2.9)–(2.10) has a positive solution wλ . Equivalently, (2.7)–
λ
(2.8) has a negative solution uλ = −wλ for λ ≥ λ̄2 . Also since ∂u
≥ 0; x ∈ ∂Ω
∂n
and f2 (S) > f (x, S); ∀x ∈ Ω̄, S ≥ 0, uλ is a strict super solution to (1.1)–(1.2) for
λ ≥ λ¯2 . Let λ̄ = max{λ¯1 , λ̄2 } and λ > λ̄ be fixed. Consider u1 (x) := Jwα (x) where
J > 0 is large enough so that
1
f (x, J||wα ||∞ )
≥
λ||wα ||∞
J||wα ||∞
(2.11)
u1 ≥ vλ .
(2.12)
and
Here (2.11) is possible since σ = 0 and (2.12) is possible by the Hopf’s maximum
∂f
principle. Then −∆u1 = J ≥ λf (x, J||wα ||∞ ) ≥ λf (x, Jwα ) (since ∂u
≥ 0) =
˜ α (x) when J˜ > 0 is large enough so that
λf (x, u1 ). Next consider v2 (x) := −Jw
˜ α ||∞ )
1
f (x, −J||w
≥
˜ α ||∞
λ||wα ||∞
−J||w
(2.13)
v2 ≤ uλ .
(2.14)
and
Here again (2.13) is possible since β = 0 and (2.14) is possible by the Hopf’s
˜ α ||∞ ) ≤ λf (x1 , −Jw
˜ α ) (since
maximum principle. Then −∆v2 = −J˜ ≤ λf (x, −J||w
∂f
∂u ≥ 0) = λf (x, v2 ). Thus u1 is a supersolution while v2 is a subsolution such that
v2 ≤ uλ ≤ 0 ≤ vλ ≤ u1 , where uλ is a strict supersolution and vλ is a strict
subsolution. Hence the result. Proof of Theorem 1.4. Let v1 (x) and u2 (x) be the strict sub and strict super
solutions to (1.1)–(1.2) as in the proof of Theorem 1.1. Then since r ≥ bmµ, both
v1 (x) and u2 (x) satisfy −r ≤ v1 (x) ≤ r, −r ≤ u2 (x) ≤ r. But by (1.16) u1 (x) = r
and v2 (x) ≡ −r are super and sub solutions respectively to (1.1)–(1.2). Hence the
result. Proof of Theorem 1.5. Let v1 (x) and u2 (x) be as in the proof of Theorem 1.4.
But λ ∈ (λ1 , λ1 + δ). Hence v1 (x) ≥ 0 while u2 (x) ≤ 0. The rest of the proof is
identical to the proof of Theorem 1.4. Proof of Theorem 1.6. Let uλ and vλ be respectively the strict super and strict
subsolutions to (1.1)–(1.2) as in the proof of Theorem 1.3. But −r2 ≤ uλ ≤ 0 ≤
vλ ≤ r1 (see [12]) while u1 (x) ≡ r are v2 (x) = −r are super and subsolutions
respectively to (1.1)–(1.2). Hence the result. EJDE–1994/04
Non-autonomous EllipticBoundary Value Problems
7
Proof of Theorem 1.7. Assume u ≥ 0, u 6≡ 0. Then
−∆u = λf (x, u)
= λ[f (x, u) − f (x, 0)] + λf (x, 0)
∂f
= λ (x, η)u + λf (x, 0)( where 0 ≤ η ≤ u)
∂u
≤ λγu + λf (x, 0)( by (1.17)).
Thus u satisfies
−∆u − λγu ≤ λf (x, 0);
x∈Ω
∂u(x)
= 0; x ∈ ∂Ω.
∂n
Multiplying (2.15) by φ1 and integrating we obtain
Z
Z
Z
−∆uφ1 dx −
λγuφ1 dx ≤
λf (x, 0)φ1 dx ≤ 0.
u(x) + α(x)
Ω
Ω
(2.15)
(2.16)
Ω
Applying Green’s second identity we obtain
Z
Z
Z
∂u
∂φ1
{−φ1
uλ1 φ1 dx −
λγuφ1 dx ≤ 0.
+u
} ds +
∂n
∂n
∂Ω
Ω
Ω
But applying the boundary conditions we see that
Z
∂u
∂φ1
{−φ1
+u
} ds = 0.
∂n
∂n
∂Ω
Thus we obtain
Z
uφ1 (λ1 − λγ) dx ≤ 0.
Ω
But u ≥ 0, φ1 > 0 in Ω and this is a contradiction if λ < λ1 /γ. Hence the result. Proof of Theorem 1.8. Assume u ≤ 0, u 6≡ 0. Proceeding as in the proof of
Theorem 1.7 we obtain
∂f
(x, z)u + λf (x, 0)
∂u
≥ λγu + λf (x, 0)( by (1.17)).
−∆u = λ
Thus u satisfies
−∆u − λγu ≥ λf (x, 0);
x∈Ω
(2.17)
∂u(x)
= 0; x ∈ ∂Ω.
(2.18)
u(x) + α(x)
∂n
R
Multiplying (2.17) by φ1 , using Ω f (x, 0)φ1 dx ≥ 0, and proceeding as in the proof
of Theorem 1.7 we obtain
Z
uφ1 (λ1 − λγ) dx ≥ 0.
Ω
8
V. Anuradha, S. Dickens, and R. Shivaji
EJDE–1994/04
But u ≤ 0 while φ1 > 0 for x ∈ Ω, thus this is a contradiction if λ < λ1 /γ. Hence
the result. Proof of Theorem 1.9. Suppose u > 0 somewhere in Ω̄. Then ∃ some Ω1 ⊆ Ω
such that u > 0 in Ω, and either
∂u(x)
u(x) = 0 or u(x) + α(x)
= 0; x ∈ ∂Ω.
(2.19)
∂n
Let λ̃1 be the principal eigenvalue and φ̃1 > 0 be a corresponding eigenfunction of
−∆ on the region Ω1 subject toRRobin conditions (1.2) on ∂Ω1 . Note that λ̃1 ≥ λ1 .
Now since f (x, 0) ≤ 0 we have Ω f (x, 0)φ̃1 dx ≤ 0. Thus following the steps in the
proof of Theorem 1.7 we obtain
Z
Z
−∆uφ̃1 dx −
λγuφ̃1 dx ≤ 0
Ω1
and then
Ω1
Z
∂u
∂ φ̃1
[−φ̃1
+u
] ds +
∂n
∂n
∂Ω1
Z
[λ̃1 − λγ]uφ̃1 dx ≤ 0.
Ω1
But if u = 0 on ∂Ω1 , then
∂u
≤ 0 on ∂Ω1 , while if u + α ∂n
= 0 on ∂Ω1 , since
φ̃1
φ̃1 + α ∂∂n
= 0, we have
φ̃1
+ u ∂∂n
= 0 on ∂Ω1 . Thus in any case (see (2.19))
∂u
∂n
∂u
−φ̃1 ∂n
φ̃1
∂u
−φ̃1 ∂n
+ u ∂∂n
≥ 0. Hence
Z
[λ̃1 − λγ]uφ̃1 dx ≤ 0
Ω1
which is a contradiction since u > 0, φ̃1 > 0 for x ∈ Ω1 while λ < λ1 /γ ≤ λ˜1 /γ.
Hence the result. Proof of Theorem 1.10. Suppose u < 0 somewhere in Ω̄. Then ∃ some Ω1 ⊆ Ω
such that u < 0 in Ω1 and either
∂u(x)
u(x) = 0 or u(x) + α(x)
(2.20)
= 0; x ∈ ∂Ω1
∂n
RLet λ̃1 and φ̃1 > 0 be as in the proof of Theorem 1.9. Now f (x, 0) ≥ 0 and hence
f (x, 0)φ̃1 dx ≥ 0. Thus following the steps in the proof of Theorem 1.8 we obtain
Ω1
Z
Z
−∆uφ̃1 dx −
λγuφ̃1 dx ≥ 0,
Ω1
and then
Z
Ω1
∂u
∂ φ̃1
[−φ̃1
+u
] ds +
∂n
∂n
∂Ω
But if u = 0 on ∂Ω1 , then
∂u
∂n
Z
[λ̃1 − λγ]uφ̃1 dx ≥ 0.
Ω1
∂u
≥ 0 on ∂Ω1 , while if u + α ∂n
= 0 on ∂Ω1 , since
φ̃1
φ̃1
∂u
φ̃1 + α ∂∂n
= 0 on ∂Ω, we have −φ̃1 ∂n
+ u ∂∂n
= 0 on ∂Ω1 . Thus in any case (see
φ̃1
∂u
(2.20)) −φ̃1 ∂n
+ u ∂∂n
≤ 0. Hence
Z
[λ̃1 − λγ]uφ̃1 dx ≥ 0
Ω1
which is a contradiction since u < 0, φ̃1 > 0 for x ∈ Ω, while λ < λ1 /γ ≤ λ̃1 /γ.
Hence the result. EJDE–1994/04
Non-autonomous EllipticBoundary Value Problems
9
Appendix 1. Let λ ∈ (λn , λn+1 ) and consider the unique solution to the boundary
value problem
−∆Zλ − λZλ = −1; x ∈ Ω
Zλ (x) + α(x)
∂Zλ (x)
= 0;
∂n
x ∈ ∂Ω.
Let φ1 > 0 satisfy (1.6)–(1.7) for k = 1. Then
Z
Z
Z
−∆Zλ φ1 dx −
Ω
−φ1 dx
λZλ φ1 dx =
Ω
Ω
which implies
Z
∂Zλ
∂φ1
{−φ1
+ Zλ
} ds +
∂n
∂n
∂Ω
Z
Z
Zλ λ1 φ1 dx −
Ω
Z
−φ1 dx .
λZλ φ1 dx =
Ω
Ω
But
Z
∂Zλ
∂φ1
{−φ1
+ Zλ
} ds =
∂n
∂n
∂Ω
Thus
Z
{α
∂Ω
∂φ1 ∂Zλ
∂Zλ ∂φ1
−α
} ds = 0 .
∂n ∂n
∂n ∂n
Z
Z
(λ1 − λ)Zλ φ1 dx =
Ω
−φ1 dx.
Ω
But λ > λ1 and φ1 > 0 for x ∈ Ω. Hence Zλ+ 6≡ 0.
References
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V. Anuradha, Department of Mathematics and Statistics, University of Arkansas
at Little Rock, Little Rock, AR 72204-1099
S. Dickens, Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762
R. Shivaji, Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762. E-mail rs1@ra.msstate.edu
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